Question

`4000 overset(0)(A)` photons is used to break the iodine molecule, then the % of energy converted to the K.E of iodine atoms if bond dissociation energy of `I_(2)` molecule is `246.5 kJ//mol`
A. `8%`
B. `12%`
C. `17%`
D. `25%`

Answer 1

Correct Answer - C
`E = (hc)/(lambda) = 3.1 eV = 297 kJ//"mole"`
`:. E = E_(0)+KE`
% energy converted to `KE = (297 - 246.5)/(297) = 17%`